Mystical field

Our point mass is losing mass at a constant rate of $\frac{r}{1000}$ = x kilograms every second. As the speed is constant it is therefore losing momentum at 5. x every second (the rate of change of momentum). Now as force is the rate of change of momentum it follows our force F is constant (rate of change in momentum is constant) and takes on the above value. F = 5. x . If the force were not to act (we must assume in this universe the laws of physics hold) it's velocity would increase due to the conservation of momentum. So the force is constant and the work done by our force = F.10.V from speed = distance.time and work done = force.displacement . This gives us an answer of 0.25J

We know that the Work done (in its simplest form i.e. parallel vectors, all quantities constants) is given by $W = F \cdot d$ where $F$ is the force applied to the object, and $d$ is the distance this object moved. But we also know $F=\frac{dp}{dt}=\frac{d\left(mv\right)}{dt}=\frac{dm}{dt}\cdot v+m\cdot\frac{dv}{dt}$ Since the velocity is not changing, we have $\frac{dv}{dt}=0$ . We also know that $\frac{dm}{dt}=0.001\frac{\text{Kg}}{\text{s}}$ and $d=v\cdot t = 50\text{ m}$ . Substituting into the work expression we get $\boxed{W = 0.25 \text{ Joules}}$

Use einstein eqn, e=mc^2, but use v instead of c

The force can be expressed as:

$F = m \frac{dv}{dt} + \frac{dm}{dt}v$ (1)

Since the point mass moves at constant velocity, the first term in (1) is zero:

$F = \frac{dm}{dt}v = rv$

Note that since $v$ is constant, the force is also constant. Work done is expressed as W (in one dimensional motion, with constant force):

$W = Fx$ (2)

For an object in constant velocity v:

$x(t) = vt$ (3)

Combining all of these:

$W = Fx = rvx$ (4)

or

$W = rt v^2 = 0.25$ J (5)

Consider initial mass of the point mass m m

In 10 seconds , mass reduce by 10 gram. As velocity is constant $d = 50 m$

$F = \frac{10^{-2 }\times 5}{10}$

$F = 5 \times 10^{-3}$

$W = work done = F \times d$

$W = 5 \times 10^{-3} \times 50$

$W = 0.25 J$

We know, in general:

$(1) F=\frac{dp}{dt}$

$(2) P=F \times v$

$(3) E=P \times T$

So, it follows, since

$\frac{dp}{dt}=-10^{-3} \times 5$

$P=-10^{-3} \times 25$

$W=-25 \times 10^{-2}=-0.25J$

So, $|W|=0.25J$

Momentum variation: dm.V Force=dm.V/t Eenergy=dm.V.d/t=dm.V^2=10E-3x5x5=0.25

We can split the rate through dimension : g/s = 10^-3 kg/s. We know that W = F X S, and we know that F = m X a (kg m/s^2). We split this dimension into kg/s X m/s which means "evaporation rate times velocity". If we know this already, then we can do this. Say the evaporation rate is Y, hence we get W = Y X V X V X t = 10^-3 X 5 X 5 X 10 = 0.25

Work = Evaporating rate x Velocity^2 x Time

r=1g/s =0.001kg/s

Therefore Work = 0.001(5^2)(10)

=0.25 J

Since we know the element $"Evaporating Rate"$ , we use a formula :

$W = V^2 \times r \times t$

where W = Work (J), V = Velocity ( $\frac{m}{s})$ , r = Evaporating Rate ( $\frac{kg}{s})$ and t (s)

By substituting the value, we get $W$ = 0,25 J.